A ball is dropped from a height of 18 m.On each rebound it rises 2/3 of the height from which it last fell. What distance has it travelled at the instant it strikes the ground for the 5th time?

Answer 1

first strike -- 18 m
2nd strike -- 2(2/3)(18) = (4/3)(18)
3rd strike -- 2(2/3)^2 (18) = 8/9 (18)
4th strike -- 2(2/3)^3 (18) = 16/27 (18)
5th strike -- 2(2/3)^4 (18) = 32/81 (18)

total = 18 + 4 terms of the GP, with a = 72/3 , r = 2/3, n = 4

sum = 18 + (72/3)((1 - 2/3)^4 )/(1 - 2/3)
= appr 75.78 m

Answer 2

Bounce on each every height from it last fell (Bn x 2/3)
B1 = 18
B2 = 12
B3 = 8
B4 = 16/3
B5 = 32/9

D(total) = H + 2(Btotal)
= 18 + 2(18 + 12 + 8 + 16/3 +32/9) = 75.89 meter.

Answer 3

Note: The answer differs slightly from the previous one due to rounding.